Bitwise operators in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a bitwise AND between two numbers.

Rust

let result = 5 [1] 3;
println!("{}", result);

Drag options to blanks, or click blank then click option'

A|

B&

C^

D<<

Attempts:

3 left

💡 Hint

Common Mistakes

Using | instead of & will perform bitwise OR, which is different.

Using ^ will perform XOR, which is not the same as AND.

The bitwise AND operator is &. It compares each bit of two numbers and returns 1 only if both bits are 1.

2fill in blank

medium

Complete the code to perform a bitwise OR between two numbers.

Rust

let result = 5 [1] 3;
println!("{}", result);

Drag options to blanks, or click blank then click option'

A^

B&

C|

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of | will perform AND, which is different.

Using ^ will perform XOR, which is not the same as OR.

The bitwise OR operator is |. It compares each bit of two numbers and returns 1 if at least one bit is 1.

3fill in blank

hard

Fix the error in the code to perform a bitwise XOR between two numbers.

Rust

let result = 5 [1] 3;
println!("{}", result);

Drag options to blanks, or click blank then click option'

A^

B|

C<<

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using & or | instead of ^ will give different results.

Using shift operators like << or >> is incorrect here.

The bitwise XOR operator is ^. It returns 1 only if the bits are different.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps numbers to their bitwise left shift by 2, only for numbers greater than 2.

Rust

let shifted = std::collections::HashMap::from_iter((1..6).filter(|&x| x [1] 2).map(|x| (x, x [2] 2)));
println!("{:?}", shifted);

Drag options to blanks, or click blank then click option'

A>

B<

C<<

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > in the filter.

Using >> instead of << for left shift.

The filter should keep numbers greater than 2, so use >. The bitwise left shift operator is <<.

5fill in blank

hard

Fill all three blanks to create a HashMap that maps the bitwise NOT of numbers to their bitwise AND with 3, only for numbers less than 5.

Rust

let result = std::collections::HashMap::from_iter((0..10).filter(|&x| x [1] 5).map(|x| (!x, x [2] [3])));
println!("{:?}", result);

Drag options to blanks, or click blank then click option'

A<

B&

C3

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < in the filter.

Using | instead of & for bitwise AND.

Not using 3 as the mask value.

Filter numbers less than 5 using <. Use bitwise AND & with 3.